Question

A baseball is popped straight up into the air and has a hang-time of 6.25s. Determine the height to which the ball rises before it reaches its peak.(Hint : the time to rise to the peak is one half the total hang time )

Answer 1

Answer:

Explanation:

At the maximum height of the ball, the velocity will be zero. This will also take half the total time, 3.125 s.  

Now use the equation...  

Δd = vf*t - 1/2at^2  

By setting the final time at the moment of the apex of the height, vf becomes 0.  

So the equation will become...  

Δd = - 1/2at^2  

The equation now resembles the change in height from the ground to the maximum height  

Δd = - 1/2at^2 = - 1/2 (-9.8 m/s^2) (3.125 s)^2 = 47.8 m

Answer 2

A baseball is popped straight up into the air and has a hang-time of 6.25sec.

We have to find the height to which the ball rise before it reaches its peak. indirectly, it asks us to find the maximum height reached by the ball.

At maximum height,

• velocity of ball becomes zero.
• hence, final velocity of the ball, v = 0
• Let the initial velocity of the ball is u.
• acceleration due to gravity, g = 9.8 m/s²

A/c to question, hang-time is 6.25 sec and it also said that time to rise to the peak was one half of the total hang time.

it means, time of flight, T = hang-time = 6.25s

we know, time of flight = 2 × time taken to reach maximum height= 2 × time taken to reach maximum height to the same reference point.

so, time taken to reach the maximum height , t = T/2 = 6.25/2 = 3.125s

so initial velocity, u = gt = 9.8 × 3.125 = 30.625 m

now using formula,

$s=ut+\frac{1}{2}at^2$

⇒ s = 30.625 × 3.125 + 1/2 × (- 9.8) × (3.125)²

[ here negative sign in g indicates that acceleration is acting just opposite of its motion ]

⇒ s = 47.85 ≈ 48 m

Therefore the height to which the ball rises before it reaches it peak is 48m.